The standard proof runs as follows: You start with the tensor algebra of the vector space in question, say V. Then you take the quotient by the ideal generated by elements of the form
v \otimes v - q(v) 1 , for all v in V
Here q is the bilinear form on V. That works for me, the problem comes when I try to see that this ideal is not so big (the ideal is proper and non trivial). The standar reference [H.B. Lawson , M.L. Michelson, Spin Geometry] says; well calm down, this is not a big deal. Take an element in the intersection of this ideal with V, then in one hand this element has the standard expresion as a sum of products of elements in the ideal with elements in the tensor algebra. On the other hand, is an element of V, so all the term of maximal degree should be zero, and then comes the obscure trick I can't figure out. He takes a weird contraction with q, and concludes by induction that the previous fact (elements of maximal degree vanish) imply that the element itself is equal to zero.
Please, if you can clarify this last part (or gave me a fresh reference where I can find another proof scheme) I'll be grateful.
I'm with you, I clearly remember that bothering "proof by miracle" in Lawson's book
Hey again, first of all i would like to thank for your quick answers.
Fosco, i share your expression, i think the proof in Lawson & Michelson is "by miracle".
Ian, as one of my algebra professor told me once; "not because you define something, it is clear that that something really exist" (please tell that to politicians). And i think that's the whole trouble here. I already knew about the proof by filtration (it can be found a few pages after the so called "proof by miracle" ). The thing i dont like about it is that you need to assume that such a thing as a Clifford algebra really exist in order to see that it has a natural filtration that induces an graded identification with the exterior algebra. But i'll forget my modals and i'll take for good that proof.
Again thanks a lot
Greeting
Raúl
For the other members in the (hopefully not vacuous) set of readers to this questions, and justo for fun i would like to rise the following question. Does anyone understand the argument given by Lawson & Michelson in their proof of the existance of Clifford algebras? If the answer is affirmative, what's the trick?
Greetings
Raúl
Sorry, I was about to explain the trick, when I discovered an error in my argument :-(
I know this question is quite old, but I wanted to share something related to this that I came across recently. I was wandering why isn't the CA defined just as the quotient with the ideal generated by u\otimes{v}-g(u,v)1, with g the bilinear non-degenerate form. Then one would obtain a commutative algebra that would reproduce exactly the symmetric bilinear structure (since u\times{v} would be just equal to g(u,v)1). Well, it happens that for dimension larger than 1, the identity is contained in the ideal, hence it's equal to the whole tensor algebra and then the quotient is just 0. In the dimension 1 cases it coincides with the CA. I obtained a simple proof that uses two non-isotropic orthogonal vectors (this is why the dimension has to be at least 2).
Marcos Arcodía: "I was wandering why isn't the CA defined just as the quotient with the ideal generated by u\otimes{v}-g(u,v)1, with g the bilinear non-degenerate form." it can be defined that way too, often the specific definition is chosen after considering who is the audience for it ... thus, if relational quotients are not commonly used by the audience then a different definition is usually given ...
Raul Alvarez I just got a reminder about this discussion and noticed the title of it: "Does anyone know a clear proof for the existence of Clifford algebras?" I think that as posted this may become too philosophical... and similar to such questions about the Real numbers, the Complex numbers and so forth ... It may be more practical to consider the various definitions and their equivalences along with the corresponding applications and their efficiency compared to other methods ...
Vesselin Gueorguiev , I think you refer to the definition using the ideal generated by u\otimes{v}+v\otimes{u}-2g(u,v)1 which is equivalent to u\otimes{u}-g(u,u)1. But if you pick u\otimes{v}-g(u,v)1 without symmetrizing the tensor product you get a trivial ideal and not the CA.
Marcos Arcodía my point was that often the specific definition is chosen after considering who is the audience for it ... thus, if relational quotients are not commonly used by the audience then a different definition is usually given ...
Vesselin Gueorguiev Oh, absolutely, I get your point (generators and relations for instance).
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